172562
At the poles, a stretched wire of a given length vibrates in unison with a tuning fork. At the equator, for same setting to produce resonance with same fork, the vibrating length of wire
1 should be decreased
2 should be increased
3 should be 3 times the original length
4 should be same
Explanation:
A Since, We know that, frequency of vibration of wire is given by- $\mathrm{f}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ $\mathrm{f} \propto \frac{1}{\ell} \sqrt{\frac{\mathrm{mg}}{\mathrm{m}}}$ At pole value of $\mathrm{g}$ is more than equator $\mathrm{g}_{\text {equator }}<\mathrm{g}_{\text {poles }}$ for some frequency setting decreases in $\mathrm{g}$ can be compensate by decreasing length only because $M$ is constant.
MHT-CET 2020
WAVES
172563
A tuning fork produces 5 beats each when in proximity to a sonometer wires of two different lengths $40 \mathrm{~cm}$ and $44 \mathrm{~cm}$. Frequency of the tuning fork is
1 $90 \mathrm{~Hz}$
2 $105 \mathrm{~Hz}$
3 $130 \mathrm{~Hz}$
4 $145 \mathrm{~Hz}$
Explanation:
B Length of first sonometer wire $l_{1}=40 \mathrm{~cm}$ and length of second sonometer wire $l_{2}=44 \mathrm{~cm}$ Since, we know the relation between frequency (f) and length of string $(l)$ is given by, $\mathrm{f} \propto \frac{1}{l}$ And the tuning fork produces 5 beats $/ \mathrm{sec}$ Now, for sonometer wire of length $40 \mathrm{~cm}$, the frequency will be $(f+5)$ and for the sonometer wire of length 44 $\mathrm{cm}$ the frequency will be $(\mathrm{f}-5)$. Therefore, $\quad \mathrm{f}_{1} l_{1}=\mathrm{f}_{2} l_{2}$ $(f+5) \times 40=(f-5) \times 44$ $40 f+200=44 f-220$ $44 f-40 f=200+220$ $4 f=420$ $f=105 \mathrm{~Hz}$
Tripura-2021
WAVES
172564
A vibrating tuning fork produces 5 beats/second with a vibrating sonometer wire of length $20 \mathrm{~cm}$. If the length of the wire is increased to $21 \mathrm{~cm}$, number of beats/second remain the same. The frequency of the tuning fork is
1 $205 \mathrm{~Hz}$
2 $200 \mathrm{~Hz}$
3 $150 \mathrm{~Hz}$
4 $100 \mathrm{~Hz}$
Explanation:
A Let frequency of the tuning fork be $=f_{t}$ Frequency of $21 \mathrm{~cm}$ sonometer $=\mathrm{f}_{\mathrm{t}}-5$ frequency of $20 \mathrm{~cm}$ sonometer $=\mathrm{f}_{\mathrm{t}}+5$ We know that, Frequency of sonometer vibration $\mathrm{f}=\frac{\mathrm{v}}{2 \mathrm{~L}}$ $\mathrm{f}_{\mathrm{t}}-5=\frac{\mathrm{v}}{2 \times 21}$ $\mathrm{f}_{\mathrm{t}}+5=\frac{\mathrm{v}}{2 \times 20}$ For same sonometer material, $v$ will be constant (given in question) Hence, $2 \times\left(f_{t}+5\right) \times 20=2\left(f_{t}-5\right) \times 21$ $20 \mathrm{f}_{\mathrm{t}}+100=21 \mathrm{f}_{\mathrm{t}}-105$ $\mathrm{f}_{\mathrm{t}}=205 \mathrm{~Hz}$
Assam CEE-2020
WAVES
172565
What should be the length of a closed pipe to produce resonance with sound wave of wavelength $62 \mathrm{~cm}$, in fundamental mode? Neglect end correction
1 $31 \mathrm{~cm}$
2 $20.6 \mathrm{~cm}$
3 $15.5 \mathrm{~cm}$
4 $46.5 \mathrm{~cm}$
Explanation:
C Given, wavelength $(\lambda)=62 \mathrm{~cm}$ We know In a closed pipe, in fundamental mode The length of the pipe $(l)=\frac{\lambda}{4}$ $\therefore \quad l=\frac{62}{4}=15.5 \mathrm{~cm}$
172562
At the poles, a stretched wire of a given length vibrates in unison with a tuning fork. At the equator, for same setting to produce resonance with same fork, the vibrating length of wire
1 should be decreased
2 should be increased
3 should be 3 times the original length
4 should be same
Explanation:
A Since, We know that, frequency of vibration of wire is given by- $\mathrm{f}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ $\mathrm{f} \propto \frac{1}{\ell} \sqrt{\frac{\mathrm{mg}}{\mathrm{m}}}$ At pole value of $\mathrm{g}$ is more than equator $\mathrm{g}_{\text {equator }}<\mathrm{g}_{\text {poles }}$ for some frequency setting decreases in $\mathrm{g}$ can be compensate by decreasing length only because $M$ is constant.
MHT-CET 2020
WAVES
172563
A tuning fork produces 5 beats each when in proximity to a sonometer wires of two different lengths $40 \mathrm{~cm}$ and $44 \mathrm{~cm}$. Frequency of the tuning fork is
1 $90 \mathrm{~Hz}$
2 $105 \mathrm{~Hz}$
3 $130 \mathrm{~Hz}$
4 $145 \mathrm{~Hz}$
Explanation:
B Length of first sonometer wire $l_{1}=40 \mathrm{~cm}$ and length of second sonometer wire $l_{2}=44 \mathrm{~cm}$ Since, we know the relation between frequency (f) and length of string $(l)$ is given by, $\mathrm{f} \propto \frac{1}{l}$ And the tuning fork produces 5 beats $/ \mathrm{sec}$ Now, for sonometer wire of length $40 \mathrm{~cm}$, the frequency will be $(f+5)$ and for the sonometer wire of length 44 $\mathrm{cm}$ the frequency will be $(\mathrm{f}-5)$. Therefore, $\quad \mathrm{f}_{1} l_{1}=\mathrm{f}_{2} l_{2}$ $(f+5) \times 40=(f-5) \times 44$ $40 f+200=44 f-220$ $44 f-40 f=200+220$ $4 f=420$ $f=105 \mathrm{~Hz}$
Tripura-2021
WAVES
172564
A vibrating tuning fork produces 5 beats/second with a vibrating sonometer wire of length $20 \mathrm{~cm}$. If the length of the wire is increased to $21 \mathrm{~cm}$, number of beats/second remain the same. The frequency of the tuning fork is
1 $205 \mathrm{~Hz}$
2 $200 \mathrm{~Hz}$
3 $150 \mathrm{~Hz}$
4 $100 \mathrm{~Hz}$
Explanation:
A Let frequency of the tuning fork be $=f_{t}$ Frequency of $21 \mathrm{~cm}$ sonometer $=\mathrm{f}_{\mathrm{t}}-5$ frequency of $20 \mathrm{~cm}$ sonometer $=\mathrm{f}_{\mathrm{t}}+5$ We know that, Frequency of sonometer vibration $\mathrm{f}=\frac{\mathrm{v}}{2 \mathrm{~L}}$ $\mathrm{f}_{\mathrm{t}}-5=\frac{\mathrm{v}}{2 \times 21}$ $\mathrm{f}_{\mathrm{t}}+5=\frac{\mathrm{v}}{2 \times 20}$ For same sonometer material, $v$ will be constant (given in question) Hence, $2 \times\left(f_{t}+5\right) \times 20=2\left(f_{t}-5\right) \times 21$ $20 \mathrm{f}_{\mathrm{t}}+100=21 \mathrm{f}_{\mathrm{t}}-105$ $\mathrm{f}_{\mathrm{t}}=205 \mathrm{~Hz}$
Assam CEE-2020
WAVES
172565
What should be the length of a closed pipe to produce resonance with sound wave of wavelength $62 \mathrm{~cm}$, in fundamental mode? Neglect end correction
1 $31 \mathrm{~cm}$
2 $20.6 \mathrm{~cm}$
3 $15.5 \mathrm{~cm}$
4 $46.5 \mathrm{~cm}$
Explanation:
C Given, wavelength $(\lambda)=62 \mathrm{~cm}$ We know In a closed pipe, in fundamental mode The length of the pipe $(l)=\frac{\lambda}{4}$ $\therefore \quad l=\frac{62}{4}=15.5 \mathrm{~cm}$
172562
At the poles, a stretched wire of a given length vibrates in unison with a tuning fork. At the equator, for same setting to produce resonance with same fork, the vibrating length of wire
1 should be decreased
2 should be increased
3 should be 3 times the original length
4 should be same
Explanation:
A Since, We know that, frequency of vibration of wire is given by- $\mathrm{f}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ $\mathrm{f} \propto \frac{1}{\ell} \sqrt{\frac{\mathrm{mg}}{\mathrm{m}}}$ At pole value of $\mathrm{g}$ is more than equator $\mathrm{g}_{\text {equator }}<\mathrm{g}_{\text {poles }}$ for some frequency setting decreases in $\mathrm{g}$ can be compensate by decreasing length only because $M$ is constant.
MHT-CET 2020
WAVES
172563
A tuning fork produces 5 beats each when in proximity to a sonometer wires of two different lengths $40 \mathrm{~cm}$ and $44 \mathrm{~cm}$. Frequency of the tuning fork is
1 $90 \mathrm{~Hz}$
2 $105 \mathrm{~Hz}$
3 $130 \mathrm{~Hz}$
4 $145 \mathrm{~Hz}$
Explanation:
B Length of first sonometer wire $l_{1}=40 \mathrm{~cm}$ and length of second sonometer wire $l_{2}=44 \mathrm{~cm}$ Since, we know the relation between frequency (f) and length of string $(l)$ is given by, $\mathrm{f} \propto \frac{1}{l}$ And the tuning fork produces 5 beats $/ \mathrm{sec}$ Now, for sonometer wire of length $40 \mathrm{~cm}$, the frequency will be $(f+5)$ and for the sonometer wire of length 44 $\mathrm{cm}$ the frequency will be $(\mathrm{f}-5)$. Therefore, $\quad \mathrm{f}_{1} l_{1}=\mathrm{f}_{2} l_{2}$ $(f+5) \times 40=(f-5) \times 44$ $40 f+200=44 f-220$ $44 f-40 f=200+220$ $4 f=420$ $f=105 \mathrm{~Hz}$
Tripura-2021
WAVES
172564
A vibrating tuning fork produces 5 beats/second with a vibrating sonometer wire of length $20 \mathrm{~cm}$. If the length of the wire is increased to $21 \mathrm{~cm}$, number of beats/second remain the same. The frequency of the tuning fork is
1 $205 \mathrm{~Hz}$
2 $200 \mathrm{~Hz}$
3 $150 \mathrm{~Hz}$
4 $100 \mathrm{~Hz}$
Explanation:
A Let frequency of the tuning fork be $=f_{t}$ Frequency of $21 \mathrm{~cm}$ sonometer $=\mathrm{f}_{\mathrm{t}}-5$ frequency of $20 \mathrm{~cm}$ sonometer $=\mathrm{f}_{\mathrm{t}}+5$ We know that, Frequency of sonometer vibration $\mathrm{f}=\frac{\mathrm{v}}{2 \mathrm{~L}}$ $\mathrm{f}_{\mathrm{t}}-5=\frac{\mathrm{v}}{2 \times 21}$ $\mathrm{f}_{\mathrm{t}}+5=\frac{\mathrm{v}}{2 \times 20}$ For same sonometer material, $v$ will be constant (given in question) Hence, $2 \times\left(f_{t}+5\right) \times 20=2\left(f_{t}-5\right) \times 21$ $20 \mathrm{f}_{\mathrm{t}}+100=21 \mathrm{f}_{\mathrm{t}}-105$ $\mathrm{f}_{\mathrm{t}}=205 \mathrm{~Hz}$
Assam CEE-2020
WAVES
172565
What should be the length of a closed pipe to produce resonance with sound wave of wavelength $62 \mathrm{~cm}$, in fundamental mode? Neglect end correction
1 $31 \mathrm{~cm}$
2 $20.6 \mathrm{~cm}$
3 $15.5 \mathrm{~cm}$
4 $46.5 \mathrm{~cm}$
Explanation:
C Given, wavelength $(\lambda)=62 \mathrm{~cm}$ We know In a closed pipe, in fundamental mode The length of the pipe $(l)=\frac{\lambda}{4}$ $\therefore \quad l=\frac{62}{4}=15.5 \mathrm{~cm}$
172562
At the poles, a stretched wire of a given length vibrates in unison with a tuning fork. At the equator, for same setting to produce resonance with same fork, the vibrating length of wire
1 should be decreased
2 should be increased
3 should be 3 times the original length
4 should be same
Explanation:
A Since, We know that, frequency of vibration of wire is given by- $\mathrm{f}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ $\mathrm{f} \propto \frac{1}{\ell} \sqrt{\frac{\mathrm{mg}}{\mathrm{m}}}$ At pole value of $\mathrm{g}$ is more than equator $\mathrm{g}_{\text {equator }}<\mathrm{g}_{\text {poles }}$ for some frequency setting decreases in $\mathrm{g}$ can be compensate by decreasing length only because $M$ is constant.
MHT-CET 2020
WAVES
172563
A tuning fork produces 5 beats each when in proximity to a sonometer wires of two different lengths $40 \mathrm{~cm}$ and $44 \mathrm{~cm}$. Frequency of the tuning fork is
1 $90 \mathrm{~Hz}$
2 $105 \mathrm{~Hz}$
3 $130 \mathrm{~Hz}$
4 $145 \mathrm{~Hz}$
Explanation:
B Length of first sonometer wire $l_{1}=40 \mathrm{~cm}$ and length of second sonometer wire $l_{2}=44 \mathrm{~cm}$ Since, we know the relation between frequency (f) and length of string $(l)$ is given by, $\mathrm{f} \propto \frac{1}{l}$ And the tuning fork produces 5 beats $/ \mathrm{sec}$ Now, for sonometer wire of length $40 \mathrm{~cm}$, the frequency will be $(f+5)$ and for the sonometer wire of length 44 $\mathrm{cm}$ the frequency will be $(\mathrm{f}-5)$. Therefore, $\quad \mathrm{f}_{1} l_{1}=\mathrm{f}_{2} l_{2}$ $(f+5) \times 40=(f-5) \times 44$ $40 f+200=44 f-220$ $44 f-40 f=200+220$ $4 f=420$ $f=105 \mathrm{~Hz}$
Tripura-2021
WAVES
172564
A vibrating tuning fork produces 5 beats/second with a vibrating sonometer wire of length $20 \mathrm{~cm}$. If the length of the wire is increased to $21 \mathrm{~cm}$, number of beats/second remain the same. The frequency of the tuning fork is
1 $205 \mathrm{~Hz}$
2 $200 \mathrm{~Hz}$
3 $150 \mathrm{~Hz}$
4 $100 \mathrm{~Hz}$
Explanation:
A Let frequency of the tuning fork be $=f_{t}$ Frequency of $21 \mathrm{~cm}$ sonometer $=\mathrm{f}_{\mathrm{t}}-5$ frequency of $20 \mathrm{~cm}$ sonometer $=\mathrm{f}_{\mathrm{t}}+5$ We know that, Frequency of sonometer vibration $\mathrm{f}=\frac{\mathrm{v}}{2 \mathrm{~L}}$ $\mathrm{f}_{\mathrm{t}}-5=\frac{\mathrm{v}}{2 \times 21}$ $\mathrm{f}_{\mathrm{t}}+5=\frac{\mathrm{v}}{2 \times 20}$ For same sonometer material, $v$ will be constant (given in question) Hence, $2 \times\left(f_{t}+5\right) \times 20=2\left(f_{t}-5\right) \times 21$ $20 \mathrm{f}_{\mathrm{t}}+100=21 \mathrm{f}_{\mathrm{t}}-105$ $\mathrm{f}_{\mathrm{t}}=205 \mathrm{~Hz}$
Assam CEE-2020
WAVES
172565
What should be the length of a closed pipe to produce resonance with sound wave of wavelength $62 \mathrm{~cm}$, in fundamental mode? Neglect end correction
1 $31 \mathrm{~cm}$
2 $20.6 \mathrm{~cm}$
3 $15.5 \mathrm{~cm}$
4 $46.5 \mathrm{~cm}$
Explanation:
C Given, wavelength $(\lambda)=62 \mathrm{~cm}$ We know In a closed pipe, in fundamental mode The length of the pipe $(l)=\frac{\lambda}{4}$ $\therefore \quad l=\frac{62}{4}=15.5 \mathrm{~cm}$
